I am really stuck in this proof. Any hint or suggestion from you would be appreciated.
Let $U \subset \Bbb{R^{n}}$ be open, bounded, with smooth boundary $\partial U$, and $ T \gt 0$. Use energy method to prove that the initial boundary-value problem
$u_{t}-\Delta u =f$ in $U_{T}$
$\frac{\partial u}{\partial \nu}=0$ on $\partial U \times [0,T]$
$u=g$ on $U \times \{t=0\}$
,where $f=f(x,t),g=g(x) $ are given smooth functions, has at most one solution $u \in C^{2}_{1}$ $(\overline U_{T})$
Best Answer
HINT: Start by assume you have two solutions $u_{1}$ and $u_{2}$ for the given problem, and show that $u=u_{1}-u_{2}$ is identically 0, because $u$ satisfies $u_{t}=\Delta u$, with $u$ having 0 boundary data. The trick to showing $u$ is the zero function is to integrate $u_{t}-\Delta u$ against $u$, for example, and use something like $ u \Delta u = \nabla\cdot (u\nabla u)-\nabla u\cdot\nabla u$. Then follow your nose using Calculus identities and use the fact that you have 0 boundary data.
So suppose $u_{1}$ and $u_{2}$ are solutions of the problem. Then $u=u_{1}-u_{2}$ satisfies $$ u_{t}-\Delta u = 0,\;\;\; \frac{\partial u}{\partial\nu}=0 \mbox{ on } \partial U \times [0,T], \;\;\; u = 0 \mbox{ on } U \times \{ t= 0\}. $$ Therefore, $$ 0=\int_{U}uu_{t}-u\Delta u\,dx = \int_{U}uu_{t}-\nabla\cdot(u\nabla u)+|\nabla u|^{2}\,dx $$ $$ = \int_{U}uu_{t}+|\nabla u|^{2}\,dx-\int_{\partial U}u\frac{\partial u}{\partial\nu}\,dS(x), $$ where $dS$ is the surface area element on $\partial U$. (This is Gauss' Divergence Theorem.) Because $\frac{\partial u}{\partial \nu}=0$ on $\partial U$ and $2uu_{t}=(u^{2})_{t}$, the above becomes $$ 0 = \int_{U}\frac{1}{2}\frac{\partial}{\partial t}u^{2}+|\nabla u|^{2}dx $$ Now integrate the above on $[0,t']$ for any $0 \le t' \le T$, and use the fact that $u(x,t)|_{t=0}=0$: $$ 0 = \int_{U}\frac{1}{2}|u(x,t')|^{2}\,dx +\int_{0}^{t'}\int_{U}|\nabla u(x,t)|^{2}dxdt. $$ Both terms on the right are non-negative and, therefore, both must be separately 0. Just considering the first term leads to the conclusion that $u=0$ for all $x \in U$ and $0 \le t' \le T$ because $u$ is continuous. Therefore $u_{1}=u_{2}$ as desired.